Is this what you mean?

 

how can i do that?

Didn't I explain?
When the (+) input of an opamp is biased with a two resistors and one capacitor voltage divider at half the supply voltage and the input and output have coupling capacitors then the output of the opamp can swing symmetrically up and down without a negative supply.

if i replace R11 with a 10k pot, will it act as the master volume control?

Not exactly.
A volume control can adjust a level from zero to maximum.
R11 can be a master gain control if it has a resistor in series to prevent the gain from being too high.

 

Your complete circuit looks pretty good.
But the extremely low output impedance of the mixer opamp shorts the AC negative feedback of the power amp resulting in very high gain and very high distortion. I added a 4.7k input resistor to the power amp to reduce its gain to 9.2 and now its distortion is low.
I reduced the value of R31 in the power amp so it gives enough base current to the NPN output transistor. My simulation does not use the transistors you will use so I changed the value of the negative feedback/bias resistor for Q1 to match my transistors.

You have two voltage dividers for the "half supply bias voltage" for the opamps, but only one is needed for both. Your diode and 100 ohm resistor do nothing and can be removed.

Your tone controls opamp is missing an output coupling capacitor to feed the volume control. Make its polarity correct.
The polarity of the capacitor from that volume control to the mixer is backwards.

The polarity of the other coupling capacitor to the mixer is also backwards.

The polarity of the coupling capacitor from the mixer opamp to the power amplifier is also backwards.

 

how about this?
how can i make the output power to 5watts?

 

The coupling capacitor at the output of the tone controls opamp now has the correct polarity.

5W into 8 ohms is (square root of 5 x 8= 6.33V RMS) which is 17.9V peak-to-peak. If the power supply is about +20V then the output will be 5W into 8 ohms.

You changed R34 to 47k so that the power amplifier has no voltage gain. Then there is no extra gain to play faint sounds louder. If R34 is 4.7k Like in my simulation then you will have an extra 20dB of volume adjustment.

 

so i need another DC source, or can i use 20V source to all other parts ?.. i changed R34 to 22k because 4.7k gives a distorted out in my simulations, is that alright?

 

You can power all the circuitry from 20VDC. Then the output of the power amp will be 5W.

If the output is clipping then simply turn down the volume control, don't reduce the gain.

 

Will you be able to explain the benefit of bootstrapping R31 in the power amplifier with the output capacitor and the speaker to your professor?

Please add these corrections:

 

actually no, what is the advantage of bootstrapping and computations?

also, i know the purpose of negative feedback---to decrease gain so as distortion..

i replaced the other transistor with 2N4401..

another question, i am thinking of using 18V source, i have no time to design for the power supply, if i use 18V, what will be the output power?

in the pre-amp, i replaced 0.47uF capacitor of 1uF because there's no available value of that from the shops i went a while ago.

 

ahhm how can i clamp the diodes to the heatsink of transistors?

 

what is the advantage of bootstrapping and computations?

The collector load resistor for the NPN driver transistor also turns on the NPN output transistor. The voltage gain of the driver transistor is fairly low with the low value collector load resistor when there is no bootstrapping. The base of the NPN output transistor can swing up to about 1.2V less than the power supply voltage and it runs out of base current which causes the horrible distortion. The maximum output power is reduced.
With bootstrapping, the upper end of the base resistor for the NPN output transistor swings up to 17V (when the supply is 12V) so it provides plenty of base current. Since the resistor has a constant voltage across it then it is a high impedance which provides more voltage gain. The maximum output power is high. The gain is high. The distortion is low.

i am thinking of using 18V source, i have no time to design for the power supply, if i use 18V, what will be the output power?

My simulation shows 4.0W. To make an unregulated 18VDC supply, you need a 14V transformer driving a full-wave bridge rectifier, or a 27V center-tapped transformer with two rectifier diodes. The transformer voltages are very rare.

in the pre-amp, i replaced 0.47uF capacitor of 1uF because there's no available value of that from the shops i went a while ago.

That is fine, it allows the low frequency response to go lower.

 

ahh see, so that's the reason for bootstrapping, to provide base current, and when there is a resistor across it, it provides a high impedance, lowering the distortion.. haha

 

From your Personal Message (I cannot attach my results):

how will i able to compute for the power of the load? i only know a simple formula for a common class AB amplifier, but this miscellaneous biasing confuses me..

Feed the input of the amplifier with a sinewave of about 1kHz.
Connect a power resistor with its value the rated load of the amplifier.
View the output signal on an oscilloscope and turn up ther volume until the amplifier is distorted or is barely clipping.
Then calculate the RMS voltage from the peak-to-peak voltage shown on the 'scope as V RMS= Vp-p/2.828. The power is V RMS squared divided by the load resistance.

I am using different output transistors so your amplifier might need its collector load resistor for the first transistor changed a little for symmetrical clipping. Your output power will be a little different from mine.

Here is my simulation with an 18V power supply:

 

i replaced the source with 24V, then the oscilloscope reads 17Vp-p, the output power is 4.5 watts. it's a good news for me because i can buy that supply voltage.. thanks again cool dude

 

With a 24VDC supply, my simulation shows 22V p-p and the output power calculates to 7.5W.

 

wow, 2n4401 is not a power transistor right? so if i us that in my circuit it will probably overheat, am i correct?

 

how can i clamp the diodes to the heatsink of transistors?

Maybe you can fiddle around with a piece of wire used as a clamp.
But if you bolt a little power transistor to the heatsink (with an insulator for it and for the output transistors) then it works exactly like two diodes. The base resistors can be adjusted to adjust the idle current in the output transistors.

 

wow, 2n4401 is not a power transistor right? so if i us that in my circuit it will probably overheat, am i correct?

With a 24V supply, the current in the 2N4401 driver transistor is about 46mA and it has a voltage of about 11.1V across it. Then its heating calculates at 46mA x 11.1V= 510mW which is close to its max allowed dissipation of 625mW at room temperature.

Yes you can use a BD135 instead that has a max allowed dissipation of 1.25W without a heatsink.

 

i found this image on google about diodes in heat sink.. is this right? the heat sink wall in the image is the heat sink of the power transistor
http://users.ece.gatech.edu/~mleach/lowtim/graphics/diodes.gif

 

"Instant bonding glue" does not conduct heat well.